Prepared by Muhammad Tayyab, Subject Specialist Mathematics, Govt Christian High School Daska
📌 Based on National Curriculum 2023 / PECTAA 2026 Syllabus
📖 What's Inside: This exercise covers addition, subtraction, multiplication, division of complex numbers, additive inverse (\( -z \)), multiplicative inverse (\( z^{-1} \)), and verification of algebraic laws (commutative, associative, etc.). Complete step-by-step solutions as per official PECTAA 2026 pattern.
📚 Related Resources – Unit 1: Complex Numbers
Class 10 Math Unit 1 Exercise 1.2 – Operations on Complex Numbers: Complete Guide
Exercise 1.2 of Unit 1 takes the foundational concepts from Exercise 1.1 and moves them into operations on complex numbers. This exercise is where students learn to add, subtract, multiply, and divide complex numbers, find additive and multiplicative inverses, and verify algebraic laws. These skills are essential for the remainder of Unit 1 and form the basis of many problems in later units.
What You Will Learn
By working through Exercise 1.2, students master the following operations: addition and subtraction of complex numbers (combining real and imaginary parts), multiplication using the distributive property and \( \iota^2 = -1 \), division using the conjugate of the denominator, finding additive inverses (\( -z \)), finding multiplicative inverses (\( z^{-1} \)), and verifying algebraic laws like commutativity and associativity.
Topics Covered in This Exercise
- Addition of Complex Numbers – adding corresponding real and imaginary parts
- Subtraction of Complex Numbers – subtracting corresponding real and imaginary parts
- Multiplication of Complex Numbers – using FOIL method with \( \iota^2 = -1 \)
- Division of Complex Numbers – multiplying numerator and denominator by conjugate
- Additive Inverse – finding \( -z = -a - b\iota \)
- Multiplicative Inverse – finding \( z^{-1} = \frac{a - b\iota}{a^2 + b^2} \)
- Verification of Laws – commutative, associative, additive identity
Why Exercise 1.2 Is Important
Exercise 1.2 is the operational core of Unit 1. Board exam papers frequently feature questions that require students to add, multiply, or divide complex numbers and find inverses. These operations appear in both short-answer and multi-part questions. Without the skills from Exercise 1.2, students cannot successfully tackle Exercise 1.3 or 1.4.
Punjab Board Preparation
Students preparing for board exams under any of the 10 BISE Punjab boards should prioritise this exercise. The division of complex numbers using conjugates is one of the most common question types, and multiplicative inverse questions appear frequently. Practising every part of Q1, Q3, and Q7 by hand is highly recommended.
PECTAA 2026 Syllabus Alignment
These solved notes are aligned with the Punjab Education & Curriculum Textbook Authority (PECTAA) 2026 syllabus, following the National Curriculum 2023 framework for Class 10 Mathematics. The question numbering, formulas, and terminology match the current Punjab textbook exactly.
Exam Tips for Complex Number Operations
- Always simplify \( \iota^2 = -1 \) immediately when multiplying.
- For division, always multiply numerator and denominator by the conjugate of the denominator.
- For multiplicative inverse, remember the formula \( z^{-1} = \frac{\bar{z}}{|z|^2} \).
- Double-check signs when subtracting complex numbers.
Common Mistakes Students Make
- Forgetting that \( \iota^2 = -1 \) when multiplying complex numbers.
- Not using the conjugate when dividing complex numbers.
- Confusing additive inverse with multiplicative inverse.
- Not simplifying final answers properly.
📖 Exercise 1.2 – Step-by-Step Solutions
1
Simplify and write in the form \(a + b\iota\)
(i) \((2 + 5\iota) + (3 - z\iota)\)
\[ \begin{aligned} (2 + 5\iota) + (3 - z\iota) &= 2 + 5\iota + 3 - z\iota \\ &= 2 + 3 + 5\iota - z\iota \\ &= 5 + (5 - z)\iota \end{aligned} \]
\[ \begin{aligned} (2 + 5\iota) + (3 - z\iota) &= 2 + 5\iota + 3 - z\iota \\ &= 2 + 3 + 5\iota - z\iota \\ &= 5 + (5 - z)\iota \end{aligned} \]
(ii) \((16 - 3\iota) + (9 + 2\iota)\)
\[ \begin{aligned} (16 - 3\iota) + (9 + 2\iota) &= 16 - 3\iota + 9 + 2\iota \\ &= 16 + 9 - 3\iota + 2\iota \\ &= 25 - \iota \end{aligned} \]
\[ \begin{aligned} (16 - 3\iota) + (9 + 2\iota) &= 16 - 3\iota + 9 + 2\iota \\ &= 16 + 9 - 3\iota + 2\iota \\ &= 25 - \iota \end{aligned} \]
(iii) \((9 - 2\iota) - (7 - 3\iota)\)
\[ \begin{aligned} (9 - 2\iota) - (7 - 3\iota) &= 9 - 2\iota - 7 + 3\iota \\ &= 9 - 7 - 2\iota + 3\iota \\ &= 2 + \iota \end{aligned} \]
\[ \begin{aligned} (9 - 2\iota) - (7 - 3\iota) &= 9 - 2\iota - 7 + 3\iota \\ &= 9 - 7 - 2\iota + 3\iota \\ &= 2 + \iota \end{aligned} \]
(iv) \((11 + 9\iota) - (9 - 7\iota)\)
\[ \begin{aligned} (11 + 9\iota) - (9 - 7\iota) &= 11 + 9\iota - 9 + 7\iota \\ &= 11 - 9 + 9\iota + 7\iota \\ &= 2 + 16\iota \end{aligned} \]
\[ \begin{aligned} (11 + 9\iota) - (9 - 7\iota) &= 11 + 9\iota - 9 + 7\iota \\ &= 11 - 9 + 9\iota + 7\iota \\ &= 2 + 16\iota \end{aligned} \]
(v) \((3 + 4\iota)(2 - 3\iota)\)
\[ \begin{aligned} (3 + 4\iota)(2 - 3\iota) &= 3(2 - 3\iota) + 4\iota(2 - 3\iota) \\ &= 6 - 9\iota + 8\iota - 12\iota^2 \\ &= 6 - \iota - 12(-1) \\ &= 6 - \iota + 12 \\ &= 18 - \iota \end{aligned} \]
\[ \begin{aligned} (3 + 4\iota)(2 - 3\iota) &= 3(2 - 3\iota) + 4\iota(2 - 3\iota) \\ &= 6 - 9\iota + 8\iota - 12\iota^2 \\ &= 6 - \iota - 12(-1) \\ &= 6 - \iota + 12 \\ &= 18 - \iota \end{aligned} \]
(vi) \((5 - 2\iota)(3 - 4\iota)\)
\[ \begin{aligned} (5 - 2\iota)(3 - 4\iota) &= 5(3 - 4\iota) - 2\iota(3 - 4\iota) \\ &= 15 - 20\iota - 6\iota + 8\iota^2 \\ &= 15 - 26\iota + 8(-1) \\ &= 15 - 26\iota - 8 \\ &= 7 - 26\iota \end{aligned} \]
\[ \begin{aligned} (5 - 2\iota)(3 - 4\iota) &= 5(3 - 4\iota) - 2\iota(3 - 4\iota) \\ &= 15 - 20\iota - 6\iota + 8\iota^2 \\ &= 15 - 26\iota + 8(-1) \\ &= 15 - 26\iota - 8 \\ &= 7 - 26\iota \end{aligned} \]
(vii) \((3 - 5\iota) \div (2 - 4\iota)\)
\[ \begin{aligned} \frac{3 - 5\iota}{2 - 4\iota} &= \frac{3 - 5\iota}{2 - 4\iota} \times \frac{2 + 4\iota}{2 + 4\iota} \\ &= \frac{3(2+4\iota) - 5\iota(2+4\iota)}{(2)^2 - (4\iota)^2} \\ &= \frac{6 + 12\iota - 10\iota - 20\iota^2}{4 - 16\iota^2} \\ &= \frac{6 + 2\iota - 20(-1)}{4 - 16(-1)} \\ &= \frac{6 + 2\iota + 20}{4 + 16} \\ &= \frac{26 + 2\iota}{20} = \frac{13}{10} + \frac{1}{10}\iota \end{aligned} \]
\[ \begin{aligned} \frac{3 - 5\iota}{2 - 4\iota} &= \frac{3 - 5\iota}{2 - 4\iota} \times \frac{2 + 4\iota}{2 + 4\iota} \\ &= \frac{3(2+4\iota) - 5\iota(2+4\iota)}{(2)^2 - (4\iota)^2} \\ &= \frac{6 + 12\iota - 10\iota - 20\iota^2}{4 - 16\iota^2} \\ &= \frac{6 + 2\iota - 20(-1)}{4 - 16(-1)} \\ &= \frac{6 + 2\iota + 20}{4 + 16} \\ &= \frac{26 + 2\iota}{20} = \frac{13}{10} + \frac{1}{10}\iota \end{aligned} \]
(viii) \((5 + 2\iota) \div (6 - 3\iota)\)
\[ \begin{aligned} \frac{5 + 2\iota}{6 - 3\iota} &= \frac{5+2\iota}{6-3\iota} \times \frac{6+3\iota}{6+3\iota} \\ &= \frac{(5+2\iota)(6+3\iota)}{36 - 9\iota^2} \\ &= \frac{30 + 15\iota + 12\iota + 6\iota^2}{36 - 9(-1)} \\ &= \frac{30 + 27\iota + 6(-1)}{36 + 9} \\ &= \frac{30 - 6 + 27\iota}{45} = \frac{24 + 27\iota}{45} = \frac{8}{15} + \frac{3}{5}\iota \end{aligned} \]
\[ \begin{aligned} \frac{5 + 2\iota}{6 - 3\iota} &= \frac{5+2\iota}{6-3\iota} \times \frac{6+3\iota}{6+3\iota} \\ &= \frac{(5+2\iota)(6+3\iota)}{36 - 9\iota^2} \\ &= \frac{30 + 15\iota + 12\iota + 6\iota^2}{36 - 9(-1)} \\ &= \frac{30 + 27\iota + 6(-1)}{36 + 9} \\ &= \frac{30 - 6 + 27\iota}{45} = \frac{24 + 27\iota}{45} = \frac{8}{15} + \frac{3}{5}\iota \end{aligned} \]
2
Write additive inverse for each complex number:
(i) \(3 + 2\iota\) → Additive inverse: \(-3 - 2\iota\)
(ii) \(4 - 3\iota\) → Additive inverse: \(-4 + 3\iota\)
(iii) \(5 - 7\iota\) → Additive inverse: \(-5 + 7\iota\)
(iv) \(-\frac{2}{3} + \frac{5}{4}\iota\) → Additive inverse: \(\frac{2}{3} - \frac{5}{4}\iota\)
3
Find multiplicative inverse for each complex number:
(i) \(4 + 5\iota\)
\[ \begin{aligned} z^{-1} &= \frac{1}{4 + 5\iota} \times \frac{4 - 5\iota}{4 - 5\iota} \\ &= \frac{4 - 5\iota}{16 - 25\iota^2} = \frac{4 - 5\iota}{16 - 25(-1)} = \frac{4 - 5\iota}{41} = \frac{4}{41} - \frac{5}{41}\iota \end{aligned} \]
\[ \begin{aligned} z^{-1} &= \frac{1}{4 + 5\iota} \times \frac{4 - 5\iota}{4 - 5\iota} \\ &= \frac{4 - 5\iota}{16 - 25\iota^2} = \frac{4 - 5\iota}{16 - 25(-1)} = \frac{4 - 5\iota}{41} = \frac{4}{41} - \frac{5}{41}\iota \end{aligned} \]
(ii) \(6 + 2\iota\)
\[ \begin{aligned} z^{-1} &= \frac{1}{6 + 2\iota} \times \frac{6 - 2\iota}{6 - 2\iota} \\ &= \frac{6 - 2\iota}{36 - 4\iota^2} = \frac{6 - 2\iota}{36 - 4(-1)} = \frac{6 - 2\iota}{40} = \frac{3}{20} - \frac{1}{20}\iota \end{aligned} \]
\[ \begin{aligned} z^{-1} &= \frac{1}{6 + 2\iota} \times \frac{6 - 2\iota}{6 - 2\iota} \\ &= \frac{6 - 2\iota}{36 - 4\iota^2} = \frac{6 - 2\iota}{36 - 4(-1)} = \frac{6 - 2\iota}{40} = \frac{3}{20} - \frac{1}{20}\iota \end{aligned} \]
(iii) \(7 - 3\iota\)
\[ \begin{aligned} z^{-1} &= \frac{1}{7 - 3\iota} \times \frac{7 + 3\iota}{7 + 3\iota} \\ &= \frac{7 + 3\iota}{49 - 9\iota^2} = \frac{7 + 3\iota}{49 - 9(-1)} = \frac{7 + 3\iota}{58} = \frac{7}{58} + \frac{3}{58}\iota \end{aligned} \]
\[ \begin{aligned} z^{-1} &= \frac{1}{7 - 3\iota} \times \frac{7 + 3\iota}{7 + 3\iota} \\ &= \frac{7 + 3\iota}{49 - 9\iota^2} = \frac{7 + 3\iota}{49 - 9(-1)} = \frac{7 + 3\iota}{58} = \frac{7}{58} + \frac{3}{58}\iota \end{aligned} \]
(iv) \(\sqrt{5} - 4\iota\)
\[ \begin{aligned} z^{-1} &= \frac{1}{\sqrt{5} - 4\iota} \times \frac{\sqrt{5} + 4\iota}{\sqrt{5} + 4\iota} \\ &= \frac{\sqrt{5} + 4\iota}{5 - 16\iota^2} = \frac{\sqrt{5} + 4\iota}{5 - 16(-1)} = \frac{\sqrt{5} + 4\iota}{21} = \frac{\sqrt{5}}{21} + \frac{4}{21}\iota \end{aligned} \]
\[ \begin{aligned} z^{-1} &= \frac{1}{\sqrt{5} - 4\iota} \times \frac{\sqrt{5} + 4\iota}{\sqrt{5} + 4\iota} \\ &= \frac{\sqrt{5} + 4\iota}{5 - 16\iota^2} = \frac{\sqrt{5} + 4\iota}{5 - 16(-1)} = \frac{\sqrt{5} + 4\iota}{21} = \frac{\sqrt{5}}{21} + \frac{4}{21}\iota \end{aligned} \]
4
If \(z_1 = 2 + 5\iota,\; z_2 = 1 - 3\iota,\; z_3 = 2 + \iota\), verify the following:
(i) \(z_1 + z_2 = z_2 + z_1\)
\[ \begin{aligned} \text{L.H.S} &= (2+5\iota) + (1-3\iota) = 3 + 2\iota \\ \text{R.H.S} &= (1-3\iota) + (2+5\iota) = 3 + 2\iota \quad \Rightarrow \text{verified} \end{aligned} \]
\[ \begin{aligned} \text{L.H.S} &= (2+5\iota) + (1-3\iota) = 3 + 2\iota \\ \text{R.H.S} &= (1-3\iota) + (2+5\iota) = 3 + 2\iota \quad \Rightarrow \text{verified} \end{aligned} \]
(ii) \(z_1 z_2 = z_2 z_1\)
\[ \begin{aligned} \text{L.H.S} &= (2+5\iota)(1-3\iota) = 2 -6\iota +5\iota -15\iota^2 = 2 - \iota +15 = 17 - \iota \\ \text{R.H.S} &= (1-3\iota)(2+5\iota) = 2 +5\iota -6\iota -15\iota^2 = 2 - \iota +15 = 17 - \iota \end{aligned} \]
\[ \begin{aligned} \text{L.H.S} &= (2+5\iota)(1-3\iota) = 2 -6\iota +5\iota -15\iota^2 = 2 - \iota +15 = 17 - \iota \\ \text{R.H.S} &= (1-3\iota)(2+5\iota) = 2 +5\iota -6\iota -15\iota^2 = 2 - \iota +15 = 17 - \iota \end{aligned} \]
(iii) \((z_1 + z_2) + z_3 = z_1 + (z_2 + z_3)\)
\[ \begin{aligned} \text{L.H.S} &= [(2+5\iota)+(1-3\iota)] + (2+\iota) = (3+2\iota)+(2+\iota) = 5 + 3\iota \\ \text{R.H.S} &= (2+5\iota) + [(1-3\iota)+(2+\iota)] = (2+5\iota)+(3-2\iota) = 5 + 3\iota \end{aligned} \]
\[ \begin{aligned} \text{L.H.S} &= [(2+5\iota)+(1-3\iota)] + (2+\iota) = (3+2\iota)+(2+\iota) = 5 + 3\iota \\ \text{R.H.S} &= (2+5\iota) + [(1-3\iota)+(2+\iota)] = (2+5\iota)+(3-2\iota) = 5 + 3\iota \end{aligned} \]
(iv) \((z_1 z_2) z_3 = z_1 (z_2 z_3)\)
\[ \begin{aligned} z_1z_2 &= 17 - \iota, \quad z_2z_3 = (1-3\iota)(2+\iota) = 2+\iota -6\iota -3\iota^2 = 2 -5\iota +3 = 5 -5\iota \\ \text{L.H.S} &= (17 - \iota)(2+\iota) = 34 + 17\iota -2\iota -\iota^2 = 34 + 15\iota +1 = 35 + 15\iota \\ \text{R.H.S} &= (2+5\iota)(5 - 5\iota) = 10 -10\iota +25\iota -25\iota^2 = 10 +15\iota +25 = 35 + 15\iota \end{aligned} \]
\[ \begin{aligned} z_1z_2 &= 17 - \iota, \quad z_2z_3 = (1-3\iota)(2+\iota) = 2+\iota -6\iota -3\iota^2 = 2 -5\iota +3 = 5 -5\iota \\ \text{L.H.S} &= (17 - \iota)(2+\iota) = 34 + 17\iota -2\iota -\iota^2 = 34 + 15\iota +1 = 35 + 15\iota \\ \text{R.H.S} &= (2+5\iota)(5 - 5\iota) = 10 -10\iota +25\iota -25\iota^2 = 10 +15\iota +25 = 35 + 15\iota \end{aligned} \]
(v) \(z_1 + (-z_1) = (-z_1) + z_1 = 0\)
\[ \begin{aligned} z_1 = 2+5\iota,\; -z_1 = -2-5\iota \\ z_1+(-z_1) = (2-2)+(5\iota-5\iota)=0, \quad (-z_1)+z_1 = 0 \end{aligned} \]
\[ \begin{aligned} z_1 = 2+5\iota,\; -z_1 = -2-5\iota \\ z_1+(-z_1) = (2-2)+(5\iota-5\iota)=0, \quad (-z_1)+z_1 = 0 \end{aligned} \]
5
If \(\displaystyle \frac{(1+\iota)^2}{2-\iota} = x + \iota y\), find \(x\) and \(y\).
\[ \begin{aligned}
\frac{(1+\iota)^2}{2-\iota} &= \frac{1 + \iota^2 + 2\iota}{2-\iota} = \frac{1 - 1 + 2\iota}{2-\iota} = \frac{2\iota}{2-\iota} \\
&= \frac{2\iota}{2-\iota} \times \frac{2+\iota}{2+\iota} = \frac{2\iota(2+\iota)}{4 - \iota^2} = \frac{4\iota + 2\iota^2}{4 - (-1)} = \frac{4\iota + 2(-1)}{5} \\
&= \frac{-2 + 4\iota}{5} = -\frac{2}{5} + \frac{4}{5}\iota
\end{aligned} \]
By equating real and imaginary parts: \(x = -\frac{2}{5},\; y = \frac{4}{5}\).
6
If \((2x + \iota y)(1 - \iota) = 4 + 2\iota\), find \(x\) and \(y\).
\[ \begin{aligned}
(2x + \iota y)(1 - \iota) &= 4 + 2\iota \\
2x + \iota y &= \frac{4+2\iota}{1-\iota} = \frac{4+2\iota}{1-\iota} \times \frac{1+\iota}{1+\iota} \\
&= \frac{(4+2\iota)(1+\iota)}{1 - \iota^2} = \frac{4+4\iota+2\iota+2\iota^2}{1 - (-1)} = \frac{4 + 6\iota + 2(-1)}{2} \\
&= \frac{4 - 2 + 6\iota}{2} = \frac{2 + 6\iota}{2} = 1 + 3\iota
\end{aligned} \]
By equating real and imaginary parts: \(2x = 1 \Rightarrow x = \frac{1}{2}\), and \(y = 3\).
7
Find the values of \(a\) and \(b\) given \((a + b\iota)(2 - \iota) = 6 + 5\iota\).
\[ \begin{aligned}
(a + b\iota)(2 - \iota) &= 2a - a\iota + 2b\iota - b\iota^2 = 2a - a\iota + 2b\iota + b \\
&= (2a + b) + (-a + 2b)\iota = 6 + 5\iota
\end{aligned} \]
By equating real and imaginary parts:
\[ \begin{cases}
2a + b = 6 \\
-a + 2b = 5
\end{cases} \]
Solving: multiply first by 2: \(4a + 2b = 12\), subtract second: \((4a+2b) - (-a+2b) = 12-5 \Rightarrow 5a = 7 \Rightarrow a = \frac{7}{5}\).
Then \(2(\frac{7}{5}) + b = 6 \Rightarrow \frac{14}{5} + b = 6 \Rightarrow b = 6 - \frac{14}{5} = \frac{30-14}{5} = \frac{16}{5}\).
Thus \(a = \frac{7}{5},\; b = \frac{16}{5}\).
📈 Key Concepts & Quick Revision
📐 Operations on Complex Numbers
- Addition: \((a+b\iota) + (c+d\iota) = (a+c) + (b+d)\iota\)
- Subtraction: \((a+b\iota) - (c+d\iota) = (a-c) + (b-d)\iota\)
- Multiplication: \((a+b\iota)(c+d\iota) = (ac-bd) + (ad+bc)\iota\)
- Division: \(\frac{a+b\iota}{c+d\iota} = \frac{(a+b\iota)(c-d\iota)}{c^2+d^2}\)
📐 Additive and Multiplicative Inverses
- Additive Inverse: \(-(a+b\iota) = -a - b\iota\)
- Multiplicative Inverse: \(z^{-1} = \frac{a-b\iota}{a^2+b^2}\)
- Conjugate: \(\bar{z} = a - b\iota\)
📝 Important MCQs for Practice
1. The additive inverse of \(3 - 2\iota\) is:
2. The multiplicative inverse of \(2 + 3\iota\) is:
3. \((3+2\iota)(3-2\iota)\) equals:
🏆 Board Exam Tips & Strategy
- Always show all steps when finding inverses — examiners award marks for intermediate steps.
- For division, write the conjugate multiplication step clearly.
- Leave answers in simplified fraction form, not decimals.
- Practice all parts of Q1 and Q3 — they appear frequently in board exams.
❓ Frequently Asked Questions
What is taught in Exercise 1.2 of Class 10 Math Unit 1?
Exercise 1.2 covers operations on complex numbers including addition, subtraction, multiplication, division, additive inverse (\( -z \)), multiplicative inverse (\( z^{-1} \)), and verification of algebraic laws like commutative and associative properties.
How many questions are there in Unit 1 Exercise 1.2?
Exercise 1.2 has 7 main questions covering simplification of complex expressions (8 parts), additive inverse (4 parts), multiplicative inverse (4 parts), verification of laws (5 parts), and solving for unknowns (3 parts).
What is the additive inverse of a complex number?
The additive inverse of a complex number \( a + b\iota \) is \( -a - b\iota \). When added together, they sum to zero: \( (a + b\iota) + (-a - b\iota) = 0 \).
What is the multiplicative inverse of a complex number?
The multiplicative inverse of a complex number \( z = a + b\iota \) is \( z^{-1} = \frac{1}{a + b\iota} = \frac{a - b\iota}{a^2 + b^2} \). When multiplied together, they equal 1.
How do you divide complex numbers?
To divide complex numbers, multiply numerator and denominator by the conjugate of the denominator: \( \frac{a+b\iota}{c+d\iota} = \frac{(a+b\iota)(c-d\iota)}{c^2+d^2} \).
Is this solution according to the PECTAA 2026 syllabus?
Yes, these solutions are prepared according to the PECTAA 2026 / National Curriculum 2023 syllabus for Class 10 Mathematics.
Is this Exercise 1.2 solution valid for all Punjab Boards?
Yes, the content follows the unified Punjab textbook and is applicable to students of all 10 BISE Punjab boards.
Are solved PDF notes available for Exercise 1.2?
Yes, a complete solved PDF for Exercise 1.2 is embedded on this page and available to download for free.
Can I download the Unit 1 Exercise 1.2 solution as a PDF?
Yes, use the Download PDF button on this page to save the complete solved Exercise 1.2 notes to your device.
Is Exercise 1.2 important for Class 10 board exams?
Yes, operations on complex numbers are frequently tested in board exams. Exercise 1.2 builds essential skills in complex number arithmetic and algebraic manipulation.
Who prepared these Class 10 Math Unit 1 notes?
These notes were prepared by Muhammad Tayyab, Subject Specialist Mathematics at Govt Christian High School Daska, for Hira Science Academy.